$$ \begin{aligned}(t-3)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}t^2-6t+9\end{aligned} $$ | |
① | Find $ \left(t-3\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ t } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(t-3\right)^2 = \color{blue}{t^2} -2 \cdot t \cdot 3 + \color{red}{3^2} = t^2-6t+9\end{aligned} $$ |